m = 199.0 g, the mass of water in the pipe.
c = 4.184 J/(g-C), the specific heat of water
Q = 2793 J, thermal energy released.
Use the formula
Q = m*c*ΔT
where
ΔT is the temperature change.
That is,
[tex](199.0 \, g)*(4.184 \, \frac{J}{g-^{o}C} )*(\Delta T \, ^{o}C) = (2793 \, J)\\\\ \Delta T = \frac{2793}{(199)(4.184)} =3.355 \, ^{o}C[/tex]
Answer: ΔT = 3.4 °C (nearest tenth)